Answer to Question #286134 in Chemistry for khryst

Question

1. Xiangling, a chef from Liyue weighed ten prawn samples from a batch of prawn delivery for her restaurant. Based on these data determine the different statistical parameter asked

[You may use the stat function or spreadsheet to calculate the values for mean and standard deviation. Just show your solutions for the values of Qexp and Gexp and your conclusion.]

Prawn 1 51.56 g Prawn 6 54.03

Prawn 2 53.01 g Prawn 7 53.90

Prawn 3 57.89 g Prawn 8 53.51

Prawn 4 52.97 g Prawn 9 52.95

Prawn 5 53.56 g Prawn 10 53.24

a. Determine an outlier using Q-test (95.0% confidence)

b. Determine an outlier using Grubb’s Test (95.0% confidence)

Answer 1

Chemistry

Question:

1. Xiangling, a chef from Liyue weighed ten prawn samples from a batch of prawn delivery for her restaurant. Based on these data determine the different statistical parameter asked

[You may use the stat function or spreadsheet to calculate the values for mean and standard deviation. Just show your solutions for the values of Qexp and Gexp and your conclusion.]

Prawn 1 51.56 g Prawn 6 54.03

Prawn 2 53.01 g Prawn 7 53.90

Prawn 3 57.89 g Prawn 8 53.51

Prawn 4 52.97 g Prawn 9 52.95

Prawn 5 53.56 g Prawn 10 53.24

a. Determine an outlier using Q-test (95.0% confidence)

b. Determine an outlier using Grubb’s Test (95.0% confidence)

Answer:

a. The possible outliers in the data are the highest and the lowest values: 57.89 and 51.56, respectively. For 95% confidence, the Dixon's Q value should not exceed 0.466. The Q values for them are:

$Q(57.89) = \frac{57.89-54.03}{57.89-51.56} = 0.610$

Thus, 57.89 is an outlier.

$Q(51.56) = \frac{52.95-51.56}{54.03-51.56} = 0.563$

0.563 is higher than 0.493 (95% confidence value for the number of values 9). Thus, 51.56 is also an outlier.

b. The same values were tested for being outliers using the Grubb's test:

With 95% confidence, the G critical value is 2.176 for this data (all points). The G value calculated for 57.89 data point was 2.583, greater than the critical value, so claimed outlier.

With 95% confidence, the G critical value is 2.110 for this data except 57.89 data point. The G value calculated for 51.56 data point is 2.240 , higher than the critical value, so 51.56 is an outlier, too.

Using both methods, the two outlier data points were identified: 51.56 and 57.89 data points.